Solution
= Solution
For each $n$, let $N_n$ be the <null set> where $f_n\ne g_n$. Their union $N$ is null. At any $x\notin N$, the two <sequences> agree term by term, so their <pointwise limits> are equal: $f(x)=g(x)$. Therefore \b[$f\sim g$]. Taking the countable union first is essential: it produces one exceptional set outside which all equalities hold simultaneously.